Calculating the Speed of a Satellite using the Law of Universal Gravitational

  • #1
guyvsdcsniper
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Homework Statement:
A satellite orbits Earth at altitudes of 5000 km. What is its speed with respect to the center of Earth? The radius of the Earth is 6380 km and its mass is 5.98 x 1024 kg.
Relevant Equations:
V=sqrt(G*me/r)
V=2pir/T
My work is attached. I did get the answer right using the formula V=√(G*me/r).

Im just confused as to why I can't use V=2pi*r/T to also get the right answer? With this equation I would assume T=86400s.

I was able to use both equations to get the correct answer for a problem relating to geosynchronous satellite. The satellite for this problem was 200km from the Earth's surface, which is significantly closer than 5000km.

Does distance play a factor and if so, how?

I appreciate any and all responses.
 

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  • #3
guyvsdcsniper
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What makes you think the period would be 24 hours?
I was just assuming that based off the geosynchronous satellite I solved. I was completely wrong. I used the value I got from my answer and solved for T and the period was 12078s.

I totally jumped the gun asking the question here without really thinking about my question. Sorry.
 
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  • #4
haruspex
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I was just assuming that based off the geosynchronous satellite I solved. I was completely wrong. I used the value I got from my answer and solved for T and the period was 12078s.

I totally jumped the gun asking the question here without really thinking about my question. Sorry.
No problem.
 
  • #5
PeroK
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I was just assuming that based off the geosynchronous satellite I solved. I was completely wrong. I used the value I got from my answer and solved for T and the period was 12078s.

I totally jumped the gun asking the question here without really thinking about my question. Sorry.
You can also get the answer without the mass of the Earth or ##G##, but using the known surface gravity. The magnitude of the force on the satellite is: $$F = \frac{GM_e m}{r^2} = \frac{mR_e^2}{r^2}\frac{GM_e}{R_e^2} = \frac{mR_e^2}{r^2}g$$ And the centripetal force is $$F_c = \frac{mv^2}{r}$$ Equating these two gives: $$v = R_e\sqrt{\frac g r} = R_e\sqrt{\frac g {R_e + h}}$$
 

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